find out the relationship between boyle's temperature and vander waals constant
Answers
Let's See...
Mathematical condition for Boyle’s temperature is....
TB = [d(PV)/dP]T
When P → 0
and we know that...
Van der Waals Equation for 1 mole gas is...
(P + a/V²)(V - b) = RT
or,
P = (RT/V - b) - a/V²
Hence...
PV = {RTV/(V - b)} - a/V
So....
TB = [d(PV)/dP]T = [RT/(V - b) - {RTV/(V - b)²} + a/V²](dV/dP)T
= [{RT(V - b) - RTV/(V - b)²} + a/V²] (dV/dP)T
= [{- RTb/(V - b)²} + a/V²] (dV/dP)T
But when...
T = TB, [d(PV)/dP]T = 0 and (dV/dP)T ≠ 0
So...
RTBb/(V - b)² = a/V²
or,
TB = (a/Rb) {(V - b)/V}² Since P → 0, V is large.
Thus...
(V - b)/V ≃ 1
Hence...
TB = a/Rb
This is relation between Boyle's temperature and Van der waals constant.
The Mathematical condition for Boyle’s temperature:
TB = [d(PV)/dP]T When P→0
According to the Van der Waals Equation for 1 mole gas,
P = (RT/V - b) - a/V²
or
(P + a/V²)(V - b) = RT
Hence, PV = {RTV/(V - b)} - a/V
Thus, TB = [d(PV)/dP]T = [RT/(V - b) - {RTV/(V - b)²} + a/V²](dV/dP)T
= [{RT(V - b) - RTV/(V - b)²} + a/V²] (dV/dP)T
=[{- RTb/(V - b)²} + a/V²] (dV/dP)T
T becomes TB; the gas equation becomes
pν = RTB
T = TB, [d(PV)/dP]T = 0 and (dV/dP)T ≠ 0
RTBb/(V - b)² = a/V²
or, TB = (a/Rb) {(V - b)/V}² Since P → 0, V is large.
(V - b)/V ≃ 1
TB = a/Rb